| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2019 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Find range where function increasing/decreasing |
| Difficulty | Moderate -0.8 This is a straightforward stationary points question requiring product rule differentiation, solving a quadratic equation, and interpreting the sign of the derivative. All steps are routine for P2 level with no conceptual challenges or novel problem-solving required. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Expands: \(y = 2x^2(x-5) = 2x^3 - 10x^2\) | B1 | May be implied by sight of \(\frac{dy}{dx} = 6x^2 - 20x\). If product rule used, award for correct \(u, u', v, v'\) stated or implied |
| \(\frac{dy}{dx} = 6x^2 - 20x\) | M1 | Differentiated reducing a power by one on one term. If product rule used, look for \(\frac{dy}{dx} = Ax(x-5) + Bx^2\) |
| Sets \(\frac{dy}{dx} = 0 \Rightarrow 6x^2 - 20x = 0 \Rightarrow x = 0, \frac{10}{3}\) oe | dM1 A1 | Sets \(\frac{dy}{dx}=0\) and solves quadratic for at least one solution. Values 0 and 3.33 imply dM1. Dependent on previous M. Dividing by \(x\) to find one solution acceptable. Allow answers just written down |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| One of \(x \leq 0\) or \(x \geq \frac{10}{3}\) | M1 | Allow \(x < 0\) or \(x > \frac{10}{3}\). Must have achieved maximum two \(x\) coordinates in (a). Do not award for inequalities using solutions from \(y=0\) or if integrated |
| Both \(x \leq 0,\ x \geq \frac{10}{3}\) | A1 | Allow \(x < 0, x > \frac{10}{3}\). Ignore joining words. Allow set notation e.g. \((-\infty, 0]\cup\left[\frac{10}{3},\infty\right)\). \(\frac{10}{3}\) need not be in lowest terms |
## Question 1:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Expands: $y = 2x^2(x-5) = 2x^3 - 10x^2$ | B1 | May be implied by sight of $\frac{dy}{dx} = 6x^2 - 20x$. If product rule used, award for correct $u, u', v, v'$ stated or implied |
| $\frac{dy}{dx} = 6x^2 - 20x$ | M1 | Differentiated reducing a power by one on one term. If product rule used, look for $\frac{dy}{dx} = Ax(x-5) + Bx^2$ |
| Sets $\frac{dy}{dx} = 0 \Rightarrow 6x^2 - 20x = 0 \Rightarrow x = 0, \frac{10}{3}$ oe | dM1 A1 | Sets $\frac{dy}{dx}=0$ and solves quadratic for at least one solution. Values 0 and 3.33 imply dM1. Dependent on previous M. Dividing by $x$ to find one solution acceptable. Allow answers just written down |
**(4 marks)**
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| One of $x \leq 0$ or $x \geq \frac{10}{3}$ | M1 | Allow $x < 0$ or $x > \frac{10}{3}$. Must have achieved maximum two $x$ coordinates in (a). Do not award for inequalities using solutions from $y=0$ or if integrated |
| Both $x \leq 0,\ x \geq \frac{10}{3}$ | A1 | Allow $x < 0, x > \frac{10}{3}$. Ignore joining words. Allow set notation e.g. $(-\infty, 0]\cup\left[\frac{10}{3},\infty\right)$. $\frac{10}{3}$ need not be in lowest terms |
**(2 marks)**
---\begin{enumerate}
\item A curve $C$ has equation $y = 2 x ^ { 2 } ( x - 5 )$\\
(a) Find, using calculus, the $x$ coordinates of the stationary points of $C$.\\
(b) Hence find the values of $x$ for which $y$ is increasing.\\
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2019 Q1 [6]}}